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Topic: Michael Addition  (Read 9694 times)

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Offline orgoclear

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Michael Addition
« on: October 30, 2009, 12:12:12 AM »
The question was given to complete the reaction

I assumed that since PhMgBr is a strong base/nucleophile so, the major product will be the 1,2 addition product i.e. Ph-CH=CH-CH(OH)-Ph but the answer indicated that the mojor product is the 1,4 product.

Can you please help me out here?

A small doubt: in 1,4 addition will the product (here) be an alcohol or an aldehyde considering the alcohol formed is an enol. Will the keto form be in more percentage or the enol form. I am asking this because the answer gives the product as
Ph-CH(Ph)-CH-CH2-OH

Offline KritikalMass

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Re: Michael Addition
« Reply #1 on: October 30, 2009, 03:33:29 AM »
If I had something like ChemDraw, I would be more than happy to help you out. But, since I use linux I am unable to help you. However, it seems that after I post someone always comes in to save the day so maybe that will also be the case in this instance.

Offline jj74

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Re: Michael Addition
« Reply #2 on: October 30, 2009, 08:10:54 AM »
Are you sure the product is right ? seems that in the formula Ph-CH(Ph)-CH-CH2-OH
 that you posted there's something wrong. Maybe an H at C2 is missing ?
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Offline movies

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Re: Michael Addition
« Reply #3 on: October 30, 2009, 11:11:55 AM »
Sounds like a typo in your book to me.  I would definitely expect the 1,2-addition product to predominate in this case.  To favor the 1,4-addition product you would need to add something like copper to the mix.  The structure error that jj74 pointed out also makes me think something is wrong in your book.

Carbonyl reduction is an occasional side reaction with Grignard reagents, but usually only when the carbonyl is very sterically hindered.  This is not something I would expect you to learn in a sophomore organic class either!

Offline orgoclear

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Re: Michael Addition
« Reply #4 on: October 31, 2009, 06:23:45 AM »
Are you sure the product is right ? seems that in the formula Ph-CH(Ph)-CH-CH2-OH
 that you posted there's something wrong. Maybe an H at C2 is missing ?


Yeah that was my typo.

But the 1,4 product is highly resonance stabilised.. Although I am in favour of 1,2 product because as you said (i) steric hindrance and (ii) the nucleophile is strong but I am not entirely able to rule out the conjugated resonance of the two benzene rings

Offline movies

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Re: Michael Addition
« Reply #5 on: October 31, 2009, 02:16:29 PM »
There is no resonance between the two aromatic rings because they are separated by a CH.

Offline orgoclear

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Re: Michael Addition
« Reply #6 on: November 03, 2009, 12:13:25 AM »
I meant the carbocation involved in the formation of the 1,4 product is a benzylic one which is highly stable.
So i expect the 1,4 product to be the major one.. Although I think the product would tautomerise back to the keto form after the 1,4 product has formed

Am i correct?

Offline orgopete

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Re: Michael Addition
« Reply #7 on: November 03, 2009, 08:55:32 AM »
Can we back up on this problem?
1,4-addition product is Ph2CHCH2CHO

1,2-addition product is PhCH=CHCH(OH)Ph

My general thinking on these reactions is if you add electron donating groups to the carbonyl, like a methyl, it will retard 1,2-addition. If you add electron donating groups to the beta-carbon, like a methyl, it will retard 1,4-addition.

For this example, I agree with movies.
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Offline movies

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Re: Michael Addition
« Reply #8 on: November 03, 2009, 05:01:38 PM »
I meant the carbocation involved in the formation of the 1,4 product is a benzylic one which is highly stable.
So i expect the 1,4 product to be the major one.. Although I think the product would tautomerise back to the keto form after the 1,4 product has formed

Am i correct?

Sorry, but there is no carbocation intermediate here!  Where are you getting that from?

You are correct that after a 1,4-addition you will get a magnesium enolate that will end up as the aldehyde but only after the acidic workup.

Offline orgopete

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Re: Michael Addition
« Reply #9 on: November 04, 2009, 11:25:49 AM »
The question was given to complete the reaction

I assumed that since PhMgBr is a strong base/nucleophile so, the major product will be the 1,2 addition product i.e. Ph-CH=CH-CH(OH)-Ph but the answer indicated that the mojor product is the 1,4 product.

Can you please help me out here?

A small doubt: in 1,4 addition will the product (here) be an alcohol or an aldehyde considering the alcohol formed is an enol. Will the keto form be in more percentage or the enol form. I am asking this because the answer gives the product as
Ph-CH(Ph)-CH-CH2-OH

While we have debated in which direction this reaction might proceed, can anyone provide a literature citation of what the product of the reaction of phenylmagnesium bromide with cinnamaldehyde is? It is clear that Ph-CH(Ph)-CH2-CH2-OH is not the product. As suggested, was something added to give the 1,4-addition? Certainly, methylmagnesium bromide adds 1,2 to cinnamaldehyde and is an org syn prep.
Author of a multi-tiered example based workbook for learning organic chemistry mechanisms.

Offline movies

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Re: Michael Addition
« Reply #10 on: November 04, 2009, 01:58:20 PM »
According to J. Am. Chem. Soc. 2004, 126, 5086–5087, they got a 98% yield of the 1,2-addition product.

Offline orgoclear

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Re: Michael Addition
« Reply #11 on: November 04, 2009, 11:48:51 PM »
Check the figure below.

Offline movies

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Re: Michael Addition
« Reply #12 on: November 07, 2009, 12:19:28 PM »
The reference above does exactly this reaction and they reported a 98% yield of the 1,2-addition product.

The resonance structure you drew is a valid one, but there is another you could draw with a carbocation at the C of the carbonyl.  Based on the principle of charge separation, the carbocation on the carbonyl carbon will be a more important resonance contributor than the structure with the + at the beta-carbon.  This really gets into hard/soft theory – have you covered any of that yet?

Offline orgoclear

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Re: Michael Addition
« Reply #13 on: November 08, 2009, 01:49:20 AM »
but there is another you could draw with a carbocation at the C of the carbonyl.  Based on the principle of charge separation, the carbocation on the carbonyl carbon will be a more important resonance contributor than the structure with the + at the beta-carbon. 

Can you please draw that one?

This really gets into hard/soft theory – have you covered any of that yet?

No, I have not.. but I am looking for a place which explains that well becasue I have heard that it is very basic in understanding reactions
If you have names of any books or sites which explain it I would be glad

Offline movies

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Re: Michael Addition
« Reply #14 on: November 09, 2009, 10:26:37 AM »
The 1,2-product is drawn below.

As for hard/soft theory it goes like this:
All nucleophiles and electrophiles can be ordered by their relative hardness.  This has to do with how diffuse their charge (partial or full charge) is on the molecule or atom.  For example, H+ would be a very hard cation because all of the charge is centered on one small atom.  In contrast, the cesium cation would be relatively soft because the atom is very large but bears the same charge.  In multi-atom systems you might also need to consider other resonance structures; for example, the allyl cation is relatively soft because the positive charge is spread out across the three carbon atoms that make up the allyl group.  Nucleophiles can be classified in the same way.  For example, alkoxides are hard nucleophiles while thiolates are relatively soft.  In the case of an enone, there are two potentially electrophilic sites: the carbonyl carbon and the beta-carbon.  The carbonyl carbon is the "harder" of the two electrophilic sites because it bears most of the positive charge based on the relative contributions of the different resonance structures.

So how all that relates to reactivity is quite simple: hard nucleophiles prefer to react with hard electrophiles, soft nucleophiles prefer to react with soft electrophiles.  In the case of this question, a hard nucleophile (organomagnesium) is used so you would predict addition to occur at the harder electrophilic site, the carbonyl carbon).  If you were to use a different organometallic reagent like an organocuprate, then you would get addition to the enone.  One explanation of this is that organocuprates are softer nucleophiles than organomagnesium compounds.  (In reality it is likely quite a bit more complicated than that, but that is the pertinent explanation for this case)

EDIT: forgot to attach the picture!
« Last Edit: November 10, 2009, 10:55:29 AM by movies »

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